Answer to Monday's Hailstone Question

On Monday the task was to produce a program that was capable of displaying the sequence of numbers produced by Lothar Collatz's simple set of rules. The sequence is sometimes known as hailstone because of the way the pattern of numbers rise and fall, as if they were a hailstone caught in a weather cloud before hitting the ground.

The question was could your program display the hailstone sequence for the number 125 and the count for how many numbers in that sequence. The answer is:

125 376 188 94 47 142 71 214 107 322
161 484 242 121 364 182 91 274 137 412
206 103 310 155 466 233 700 350 175 526
263 790 395 1186 593 1780 890 445 1336 668
334 167 502 251 754 377 1132 566 283 850
425 1276 638 319 958 479 1438 719 2158 1079
3238 1619 4858 2429 7288 3644 1822 911 2734 1367
4102 2051 6154 3077 9232 4616 2308 1154 577 1732
866 433 1300 650 325 976 488 244 122 61
184 92 46 23 70 35 106 53 160 80
40 20 10 5 16 8 4 2 1
There were 109 numbers in the sequence.

Here's my version of the program:

public class Hailstone {

  public static void main(String[] args) {

    int number = 125;
    int calculation = 1;
    System.out.print(number + " ");
    while (number > 1)
    {

      if (number % 2 == 0)
      {
         //even number calculation
         number /= 2;
       }
       else
       {
         //odd number calculation
         number = (number*3) + 1;
       }
       calculation++;

       //display nicely by wrapping around every 10th calculation
       if (calculation % 10 == 0)
       {
         System.out.println(number);
       }
       else
       {
         System.out.print(number + " ");
       }

    }
    System.out.println("\nThere were " + calculation + " numbers in the sequence.");
  }
}